Linear-2 7

\begin{matrix} T : V \to V is injective \\ ⟨, ⟩_{T} : V \times V \to F \\ Show that ⟨, ⟩_{T} is an inner product \\ Solution: \\ ⟨ v + u, w ⟩_{T} = ⟨ T (v + u), T (w) ⟩ = ⟨ T (v) + T (u), T (w) ⟩ = ⟨ T (v), T (w) ⟩ + ⟨ T (u), T (w) ⟩ \\ ⟨ α v, u ⟩_{T} = ⟨ T (α v), T (u) ⟩ = ⟨ α T (v), T (u) ⟩ = α ⟨ T (v), T (u) ⟩ = α ⟨ v, u ⟩_{T} \\ ⟨ v, u ⟩_{T} = ⟨ T (v), T (u) ⟩ = \overset{―}{⟨ T (u), T (v) ⟩} = \overset{―}{⟨ u, v ⟩_{T}} \\ ⟨ v, v ⟩_{T} = ⟨ T (v), T (v) ⟩ \geq 0 \\ ⟨ v, v ⟩_{T} = 0 ⟺ ⟨ T (v), T (v) ⟩ = 0 ⟺ T (v) = 0 ⟺ v \in k e r (T) \\ \underset{―}{T is injective} ⟺ k e r (T) = {0} ⟹ ⟨, ⟩_{T} is an inner product \end{matrix}

\begin{matrix} Let V be an inner product space \\ Let {v_{1}, \dots, v_{n}} \subseteq V \\ Prove or disprove: \sum_{i = 1}^{n} \sum_{j = 1}^{n} ⟨ v_{i}, v_{j} ⟩ \geq 0 \\ Proof: \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} ⟨ v_{i}, v_{j} ⟩ = \sum_{j = 1}^{n} \sum_{i = 1}^{n} ⟨ v_{i}, v_{j} ⟩ = \sum_{j = 1}^{n} ⟨ \sum_{i = 1}^{n} v_{i}, v_{j} ⟩ = \\ = \sum_{j = 1}^{n} \overset{―}{⟨ v_{j}, \sum_{i = 1}^{n} v_{i} ⟩} = \overset{―}{\sum_{j = 1}^{n} ⟨ v_{j}, \sum_{i = 1}^{n} v_{i} ⟩} = \overset{―}{⟨ \sum_{j = 1}^{n} v_{j}, \sum_{i = 1}^{n} v_{i} ⟩} = ⟨ v, v ⟩ \geq 0 \end{matrix}

\begin{matrix} Prove: v = 0 ⟺ \forall u \in V : ⟨ v, u ⟩ = 0 \\ Let B be a basis of V \\ Prove: \forall i \in [1, n] : ⟨ v, v_{i} ⟩ = ⟨ u, v_{i} ⟩ ⟺ v = u \\ Proof: \\ Let u \in V \\ v = 0 ⟹ ⟨ v, u ⟩ = ⟨ 2 v, u ⟩ = ⟨ v, u ⟩ + ⟨ v, u ⟩ ⟹ ⟨ v, u ⟩ = 0 \\ \forall u \in V : ⟨ v, u ⟩ = 0 \underset{u = v}{⟹} ⟨ v, v ⟩ = 0 ⟹ v = 0 \\ Proof: \\ ⟨ v, v_{i} ⟩ = ⟨ u, v_{i} ⟩ ⟹ ⟨ v - u, v_{i} ⟩ = 0 \\ Let w \in V \\ w = \sum_{i = 1}^{n} α_{i} v_{i} \\ ⟨ v - u, w ⟩ = ⟨ v - u, \sum_{i = 1}^{n} α_{i} v_{i} ⟩ = \overset{―}{⟨ \sum_{i = 1}^{n} α_{i} v_{i}, v - u ⟩} = \sum_{i = 1}^{n} \overset{―}{α_{i}} \overset{―}{⟨ v - u, v_{i} ⟩} = 0 \end{matrix}

Gram-Schmidt matrix

\begin{matrix} Let S = {v_{1}, \dots, v_{n}} \subseteq V \\ G_{S} \in F^{n \times n} : (G_{S})_{i j} = ⟨ v_{i}, v_{j} ⟩ \end{matrix}

\begin{matrix} Prove: G_{S} is non-invertible ⟺ S is a linear dependence \\ Proof: \\ Let G_{S} be non-invertible \\ ⟹ \exists \sum_{j = 1}^{n} α_{j} C_{j} (G_{S}) = 0 : \exists j \in [1, n] : α_{j} \neq 0 \\ ⟹ 0 = \sum_{j = 1}^{n} α_{j} C_{j} (G_{S}) = \sum_{j = 1}^{n} α_{j} (\begin{matrix} ⟨ v_{1}, v_{j} ⟩ \\ ⟨ v_{2}, v_{j} ⟩ \\ ⋮ \\ ⟨ v_{n}, v_{j} ⟩ \end{matrix}) \\ \forall i \in [1, n] : ⟹ \sum_{j = 1}^{n} α_{j} ⟨ v_{i}, v_{j} ⟩ = 0 \\ ⟹ \sum_{j = 1}^{n} ⟨ v_{i}, \overset{―}{α_{j}} v_{j} ⟩ = 0 \\ ⟹ \sum_{j = 1}^{n} \overset{―}{⟨ \overset{―}{α_{j}} v_{j}, v_{i} ⟩} = \overset{―}{\sum_{j = 1}^{n} ⟨ \overset{―}{α_{j}} v_{j}, v_{i} ⟩} = 0 \\ ⟹ ⟨ \underset{u}{\underset{⏟}{\sum_{j = 1}^{n} \overset{―}{α_{j}} v_{j}}}, v_{i} ⟩ = 0 \\ ⟨ u, u ⟩ = ⟨ \sum_{j = 1}^{n} \overset{―}{α_{j}} v_{j}, u ⟩ = \sum_{j = 1}^{n} \overset{―}{α_{j}} ⟨ v_{j}, u ⟩ = 0 \\ ⟹ u = 0 ⟹ \sum_{i = 1}^{n} \overset{―}{α_{i}} v_{i} = 0 \\ \exists j \in [1, n] : α_{j} \neq 0 ⟹ \overset{―}{α_{j}} \neq 0 \\ ⟹ S is a linear dependence \end{matrix}